A projective morphism is proper

One of the standard facts in algebraic geometry is that a projective scheme is proper. In the language of varieties, one says that the image of a projective variety is closed. The precise statement one proves is that:

Theorem 1 Let be any variety over the algebraically closed field . Let be a closed subset. Then the projection of to is closed.

This statement is sometimes phrased as saying that is “complete.” In many ways, it is a compactness statement. Recall that a compact space has the property that the map for any is a closed map. The converse is also true if there are reasonable assumptions (I think locally compact Hausdorff on will do it). Of course, these reasonable assumptions don’t apply to a variety with the Zariski topology, but they do if we are working with a variety (say, a quasiprojective variety) over and we can define the complex topology. And in fact, it turns out that projective varieties are indeed compact in the complex topology. The scheme-theoretic version is a little different. First, the theorem for varieties said that the object was “complete” in a certain sense. But the philosophy of Grothendieck is to consider not so much schemes but morphisms of schemes, and to do everything in a relative context. The idea is:

Definition 2 A morphism is proper if it is separated, of finite type, and universally closed (i.e. any base change is a closed morphism).

I don’t really want to define separated in this post. On the other hand, I should explain what’s going on for quasiprojective varieties over an algebraically closed fields. In this case, the conditions of finite type and separated are redundant. The key condition is universal closedness, which won’t be satisfied for a general morphism. For instance, to say that is universally closed implies the first lemma about closedness of the projection.

Anyway, the scheme-theoretic version of the completeness of projective space is:

Theorem 3 The map is proper for any .

Here is defined in a purely algebraic way, as the “” of the graded algebra . I will assume that readers either know what this means or will look it up in Hartshorne or elsewhere. It is fairly easy to show directly that is separated (i.e., separated over ). In fact, more generally, the of any ring is a separated scheme. The reason is that a scheme is separated if and only if there is an open affine cover of such that each intersection is affine and generate as a ring; this is what you get when you translate the separatedness condition

For the of a graded ring, we can just use basic open affines as the . The hard part is the universal closedness. One way is to use the highly useful valuative criterion, as in Hartshorne. This has the benefit of also giving separatedness—though it doesn’t tell you that the of any graded ring is separated. Nonetheless, there is an elementary way to prove this directly as well. I want to explain that.

The statement for varieties

To approach this, let us start by reviewing the proof for varieties. In particular, we shall prove lemma 1, which goes under the name “elimination theory.” The first standard reduction is to observe that being closed is a local property. We can cover be an open affine cover, so we can just assume that is affine. Also, since is imbedded in , it suffices to show that the projection

is a closed map. So let be a closed subset. This means that the set is defined by some equations

where . Here each is homogeneous in . To say that the image is closed is the same as saying as the set of such that

has no nontrivial root is open in . But by the homogeneous Nullstellensatz, this implies that the generate a power of the irrelevant ideal in . In particular, consider the ideal . Then consider the base-change

For simplicity, we shall denote , and to be the ideal in vanishing at . We have then that for some . This is precisely the statement that the generate a power of the irrelevant ideal. In particular, if denotes the set of homogeneous polynomials of -degree in , we have

This is the same thing as saying that

The fact that is a finite module over the ring implies by Nakayama that the localized module is zero. In particular, there is but in such that . What does this buy us? Well, it means that on the open set where is invertible, we have that is just . So can have no points lying over . This is the open neighborhood of not containing anything in the image of that we wanted.

0.2. The scheme case

OK. Well, it turns out that pretty much that entire proof carries over to the case of schemes. In particular, we can show:

Proposition 4 Suppose is a commutative ring. Then the map

is a closed map.

This is enough to imply properness of (and consequently that of as properness is preserved under base change). The reason is that to show that is closed for any scheme , it reduces by covering by open affines to the case of affine itself. To prove this, let’s try to imitate the previous proof. Consider a closed subscheme . One way to get such a subscheme is to consider a homogeneous ideal and consider the inclusion

Then the closed subset is just the set of homogeneous prime ideals in containing (provided you think of via homogeneous primes). So let’s suppose that a point in , represented by a prime , has trivial fiber over . We want to show that there is an open neighborhood of which has trivial fiber. Well, first of all, we can make this more algebraic. To say that is not in the image of is to say that , where is the quotient field at and we have the standard morphism

However, the process of forming ‘s plays nice with base-change. The precise statement is that if is a morphism of rings, then the map of graded rings induces a cartesian diagram between the appropriate Proj’s and Spec’s.

Probably the easiest way to see this is to use the explicit standard open cover of by open affines. You can look this up in EGA. Anyway, back to the main proof. So we know that the fiber of over is , and similarly the fiber of is . To say that the proj of something is zero is to say that it vanishes in high degrees. This means that

is surjective for large . (Here the subscript means degree .) By Nakayama’s lemma and the finite generation of , we find that . There is thus such that

I claim that the image of doesn’t intersect . Indeed, to see this, we consider the fibered product ; this is . But the ideal contains all the homogeneous elements of degree . Since it is an ideal, it contains all homogeneous elements of degree . So

is zero in sufficiently high dimensions. Thus it has trivial . And the fiber over is empty, completing the proof.

Anyway, what’s the moral of all this? One is that properness and separatedness don’t require noetherian hypotheses to work, and the “non-elementary” means of valuative criteria is unnecessary to prove the properness of projective morphisms.

Incidentally, properness is not synonymous with projectivity. Under reasonable hypotheses, proper morphisms can be approximated by projective morphisms, though; this is called Chow’s lemma, and is crucial for proving things like the direct image theorem (the push-forward of a coherent sheaf by a proper map is coherent–ditto for the higher direct images).

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2 Responses to “A projective morphism is proper”

Akhil, are you are making the assumption that Proj S is empty precisely when the graded ring S vanishes at high degrees? This isn’t quite the case I think. Indeed, consider a graded ring S with nontrivial elements of arbitrarily high odd degrees and then kill all even degree parts. Then any element in S+ is nilpotent. But this means that S+ is contained in every prime. In particular Proj S is empty, but S doesn’t vanish at high degree.

I don’t think so — the proof only deals with graded algebras finitely generated over their degree zero component, in which case we do have an equivalence (if everything in is nilpotent, then is zero in sufficiently high degrees).